`lim(t to 0) (sqrt(cos t)- root(3)(cos t))/(sin^(2)t)`=

Options

A. `1/3`
B. `1/6`
C. `-1/12`
D. `2/3`

Answer

To solve the limit

\[ \lim_{t \to 0} \frac{\sqrt{\cos t} - \sqrt[3]{\cos t}}{\sin^2 t}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( t = 0 \) into the expression. We find:

\[ \sqrt{\cos(0)} - \sqrt[3]{\cos(0)} = \sqrt{1} - \sqrt[3]{1} = 1 - 1 = 0, \]

and

\[ \sin^2(0) = 0. \]

Thus, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

### Step 3: Differentiate the numerator and denominator
The numerator is

\[ \sqrt{\cos t} - \sqrt[3]{\cos t}. \]

We differentiate it:

1. The derivative of \( \sqrt{\cos t} = \cos^{1/2} t \) is

\[ \frac{1}{2} \cos^{-1/2} t \cdot (-\sin t) = -\frac{\sin t}{2\sqrt{\cos t}}. \]

2. The derivative of \( \sqrt[3]{\cos t} = \cos^{1/3} t \) is

\[ \frac{1}{3} \cos^{-2/3} t \cdot (-\sin t) = -\frac{\sin t}{3\sqrt[3]{\cos^2 t}}. \]

Thus, the derivative of the numerator is:

\[ -\frac{\sin t}{2\sqrt{\cos t}} + \frac{\sin t}{3\sqrt[3]{\cos^2 t}}. \]

The denominator is \( \sin^2 t \), and its derivative is

\[ 2\sin t \cos t. \]

### Step 4: Rewrite the limit using derivatives
Now we rewrite the limit using the derivatives:

\[ \lim_{t \to 0} \frac{-\frac{\sin t}{2\sqrt{\cos t}} + \frac{\sin t}{3\sqrt[3]{\cos^2 t}}}{2\sin t \cos t}. \]

### Step 5: Simplify the limit
We can factor out \( \sin t \) from the numerator:

\[ \lim_{t \to 0} \frac{\sin t \left(-\frac{1}{2\sqrt{\cos t}} + \frac{1}{3\sqrt[3]{\cos^2 t}}\right)}{2\sin t \cos t}. \]

Cancelling \( \sin t \) (valid since \( \sin t \neq 0 \) for \( t \neq 0 \)) gives:

\[ \lim_{t \to 0} \frac{-\frac{1}{2\sqrt{\cos t}} + \frac{1}{3\sqrt[3]{\cos^2 t}}}{2\cos t}. \]

### Step 6: Evaluate the limit
Now we substitute \( t = 0 \):

1. \( \cos(0) = 1 \)
2. \( \sqrt{\cos(0)} = 1 \)
3. \( \sqrt[3]{\cos^2(0)} = 1 \)

Thus, we have:

\[ \frac{-\frac{1}{2} + \frac{1}{3}}{2} = \frac{-\frac{3}{6} + \frac{2}{6}}{2} = \frac{-\frac{1}{6}}{2} = -\frac{1}{12}. \]

### Final Result
Therefore, the limit is

\[ \boxed{-\frac{1}{12}}. \]